I have come across a question while studing for my exams prove $$\log_2 x < x \text{ when }x>0$$
I know I have to solve it using a base case eg when $x=1$ then assume a inductive step $x=k$ is true but I'm having difficulty trying to solve when $x=K+1$ Can anyone help?
On
$$ \log_2(k+1) < \log_2(2k) = \log_2 2 + \log_2 k = 1+\log_2 k < 1 + k. $$
The first strict inequality holds whenever $k+1<2k$, and that happens whenever $1<k$. So prove the result when $k=1$ or $k=2$ and go on from there.
You should not use both the lower-case $k$ and the capital $K$ interchangeably in mathematical notation. Those should be treated as if they were two separate letters. That is standard usage.
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Here's how I would do this problem: Base case: If $x = 1$, then $$\log_2(x)=\dfrac{\log(x)}{\log (2)} = 0\lt 1$$
Assume when $x = k\ge 1$ we have $\log(k)/\log (2) \lt k$. Then $$\dfrac{\log(k + 1)}{\log(2)} \lt (k + 1) \dfrac{\log(k + 1)}{\log(2)} \lt \dfrac{\log(2^{k+1})}{ \log (2)}$$
This proves the statement because if the logs were removed you would get $k + 1 \lt 2^{k + 1}$ which you know is true because the base case is true, and as the value on the left side increases linearly, the value on the right side increases exponentially.
Hint. Show that $\log(k+1) - \log(k) < (k+1) - k$.